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11 Aug 2003, 07:42
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There are 20 workers in a firm. Five of them are to be sent on a mission. How many different groups could be formed, providing president, VP and accountant of the firm can't go all together among those five people?
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11 Aug 2003, 23:52
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Konstantin Lynov
Close, but not quite.
Here is an approach:
There first part of the problem is done correctly by Mciato. We do need a total number of combinations and we need to decrease it by the number of groups where the President (P), Vice President (VP) and Accountant (A) serve together (a limiting condition).
The number of the teams that VP and P would serve together on is perhaps the hardest thing in this problem. Anyway, the trick is to count on how many teams P, VP and A will be. To do this, we need to imagine the group, and the five places in it: let’s assume that P is place #1 (since the order does not really matter), VP is # 2, A is # 3, and the two other places are available to the rest (17 total), so for the for the 4th place we can have 17 candidates, and the 5th place will be offered only to the remaining 16.
Thus, the total number of groups that P, VP and A would meet is 17C2 = 16*17/2*1=136.
Answer: 20C5-17C2=15504-136=15368
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